JEE Main 2022 — Functions Question with Solution

Given one-one function

$$f:\{a,b,c,d\}\to \{0,1,2,....10\}$$

and $$2f(a)-f(b)+3f(c)+f(d)=0$$

$$\Rightarrow 3f(c)+2f(a)+f(d)=f(b)$$

Case I:

(1) Now let $$f(c)=0$$ and $$f(a)=1$$ then

$$3\times 0+2\times 1+f(d)=f(b)$$

$$\Rightarrow 2+f(d)=f(b)$$

Now possible value of $$f(d)=2,3,4,5,6,7,$$ and $$8$$.

f(d) can't be 9 and 10 as if $$f(d)=9$$ or 10 then $$f(b)=2+9=11$$ or $$f(b)=2+10=12$$, which is not possible as here any function's maximum value can be 10.

$$\therefore$$ Total possible functions when $$f(c)=0$$ and $$f(a)=1$$ are = 7

(2) When $$f(c)=0$$ and $$f(a)=2$$ then

$$3\times 0+2\times 2+f(d)=f(b)$$

$$\Rightarrow 4+f(d)=f(b)$$

$$\therefore$$ possible value of $$f(d)=1,3,4,5,6$$

$$\therefore$$ Total possible functions in this case = 5

(3) When $$f(c)=0$$ and $$f(a)=3$$ then

$$3\times 0+2\times 3+f(d)=f(b)$$

$$\Rightarrow 6+f(d)=f(b)$$

$$\therefore$$ Possible value of $$f(d)=1,2,4$$

$$\therefore$$ Total possible functions in this case = 3

(4) When $$f(c)=0$$ and $$f(a)=4$$ then

$$3\times 0+2\times 4+f(d)=f(b)$$

$$\Rightarrow 8+f(d)=f(b)$$

$$\therefore$$ Possible value of $$f(d)=1,2$$

$$\therefore$$ Total possible functions in this case = 2

(5) When $$f(c)=0$$ and $$f(a)=5$$ then

$$3\times 0+2\times 5+f(d)=f(b)$$

$$\Rightarrow 10+f(d)=f(b)$$

Possible value of f(d) can be 0 but f(c) is already zero. So, no value to f(d) can satisfy.

$$\therefore$$ No function is possible in this case.

$$\therefore$$ Total possible functions when $$f(c)=0$$ and $$f(a)=1,2,3$$ and $$4$$ are $$=7+5+3+2=17$$

Case II:

(1) When $$f(c)=1$$ and $$f(a)=0$$ then

$$3\times 1+2\times 0+f(d)=f(b)$$

$$\Rightarrow 3+f(d)=f(b)$$

$$\therefore$$ Possible value of $$f(d)=2,3,4,5,6,7$$

$$\therefore$$ Total possible functions in this case = 6

(2) When $$f(c)=1$$ and $$f(a)=2$$ then

$$3\times 1+2\times 2+f(d)=f(b)$$

$$\Rightarrow 7+f(d)=f(b)$$

$$\therefore$$ Possible value of $$f(d)=0,3$$

$$\therefore$$ Total possible functions in this case = 2

(3) When $$f(c)=1$$ and $$f(a)=3$$ then

$$3\times 1+2\times 3+f(d)=f(b)$$

$$\Rightarrow 9+f(d)=f(b)$$

$$\therefore$$ Possible value of $$f(d)=0$$

$$\therefore$$ Total possible functions in this case = 1

$$\therefore$$ Total possible functions when $$f(c)=1$$ and $$f(a)=0,2$$ and $$3$$ are

$$=6+2+1=9$$

Case III:

(1) When $$f(c)=2$$ and $$f(a)=0$$ then

$$3\times 2+2\times 0+f(d)=f(b)$$

$$\Rightarrow 6+f(d)=f(b)$$

$$\therefore$$ Possible values of $$f(d)=1,3,4$$

$$\therefore$$ Total possible functions in this case = 3

(2) When $$f(c)=2$$ and $$f(a)=1$$ then,

$$3\times 2+2\times 1+f(d)=f(b)$$

$$\Rightarrow 8+f(d)=f(b)$$

$$\therefore$$ Possible values of $$f(d)=0$$

$$\therefore$$ Total possible function in this case = 1

$$\therefore$$ Total possible functions when $$f(c)=2$$ and $$f(a)=0,1$$ are

$$=3+1=4$$

Case IV:

(1) When $$f(c)=3$$ and $$f(a)=0$$ then

$$3\times 3+2\times 0+f(d)=f(b)$$

$$\Rightarrow 9+f(d)=f(b)$$

$$\therefore$$ Possible values of $$f(d)=1$$

$$\therefore$$ Total one-one functions from four cases

$$=17+9+4+1=31$$